Rules for Calculating Probabilities: Intersection, Union, and Conditional Probability Guide

Foundations of Probability and Set Operations

• Event as a Set: Any event in a sample space $S$ is a subset of that sample space. The elements of the sample space are called outcomes, and the elements of an event are also outcomes.

• Set Theory Operations: New events are generated using operations from set theory:

  • Operation of Complement: Applied to exactly one set.

  • Operation of Intersection: Applied to two or more sets.

  • Operation of Union: Applied to two or more sets.

• Combination of Events: By applying intersection or union to two events (e.g., events $A$ and $B$), we combine them to form a new event in the same sample space.

The Intersection of Two Events

• Definition: The intersection of two events $A$ and $B$ in sample space $S$ is the event containing outcomes common to both $A$ and $B$.

• Notation: Denoted as $A \cap B$, read as "$A$ intersection $B$".

• Venn Diagram representation: Represented as a rectangle for the sample space and circles for events. The intersection is the shaded region common to both circles.

• Occurrence: Both events $A$ and $B$ must occur together in the random experiment for the intersection to occur.

• Example 1 (Tossing a coin three times):

  • Sample Space $S = \{hhh, thh, hth, hht, htt, tht, tth, ttt\}$. Total outcomes $n(S) = 8$.

  • Event $B = \{thh, hth, hht\}$ ("two heads").

  • Event $F = \{hhh, hth, hht, htt\}$ ("one head in first toss").

  • Intersection $B \cap F = \{hth, hht\}$.

  • Probability of intersection (Uniform Sample Space):         $P(B \cap F) = \frac{n(B \cap F)}{n(S)} = \frac{2}{8} = \frac{1}{4} = 0.2500$

Mutually Exclusive Events

• Definition: Two events $A$ and $B$ are mutually exclusive (or disjoint) if they have no common outcomes.

• Set Notation: $A \cap B = \emptyset$ (where $\emptyset$ is the empty set $\emptyset = \{\}$).

• Probability Property: The probability of the intersection of mutually exclusive events is always zero: $P(A \cap B) = 0$.

• Implication: Mutually exclusive events cannot occur together in the same random experiment.

• Simple Events: Events consisting of only one outcome are always mutually exclusive.

• Example 1 (Cont.):

  • Event $A = \{htt, tht, tth\}$ ("one head").

  • Event $B = \{thh, hth, hht\}$ ("two heads").

  • Since they have no common outcomes, $A \cap B = \emptyset$ and $P(A \cap B) = 0$.

The Union of Two Events

• Definition: The union of events $A$ and $B$ in sample space $S$ is the event whose outcomes are in $A$, or in $B$, or in both.

• Notation: Denoted as $A \cup B$, read as "$A$ union $B$".

• Venn Diagram representation: The shaded region belonging to at least one of the two circles.

• Listing Outcomes: When listing outcomes in a union, each outcome is listed only once, even if it belongs to both events.

• Example 1 (Cont.):

  • $B = \{thh, hth, hht\}$.

  • $F = \{hhh, hht, hth, htt\}$.

  • $B \cup F = \{thh, hth, hht, hhh, htt\}$.

  • Probability (Uniform Sample Space):         $P(B \cup F) = \frac{n(B \cup F)}{n(S)} = \frac{5}{8} = 0.6250$

Complement of an Event

• Definition: The complement of event $A$ contains all outcomes in the sample space $S$ that do not belong to $A$.

• Notation: Denoted as $A^c$ (also commonly seen as $\bar{A}$ or $A'$).

• Properties:

  • The union of an event and its complement equals the entire sample space: $A \cup A^c = S$.

  • An event and its complement are always mutually exclusive: $A \cap A^c = \emptyset$.

  • The probability of their intersection is zero: $P(A \cap A^c) = 0$.

• Example 1 (Cont.):

  • Event $G = \{hhh, thh, hth, hht, htt, tht, tth\}$ ("at least one head").

  • Complement $G^c = \{ttt\}$ ("no heads").

Addition Rules for Calculating Probabilities

General Addition Rule

• Used for any two events $A$ and $B$ in a sample space (uniform or non-uniform) that are not necessarily mutually exclusive.

• Formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

• Explanation: The subtraction of the intersection probability is necessary because $P(A \cap B)$ is counted twice when adding $P(A)$ and $P(B)$.

Special Addition Rule (Mutually Exclusive Events)

• If events $A$ and $B$ are mutually exclusive, then $P(A \cap B) = 0$.

• Formula: $P(A \cup B) = P(A) + P(B)$.

Rule for Complements

• Used to find the probability of the complement if the probability of the event is known.

• Formula: $P(A^c) = 1 - P(A)$.

• Proof Summary: Since $A \cup A^c = S$, $P(A \cup A^c) = P(S) = 1$. Because they are mutually exclusive, $P(A) + P(A^c) = 1$, leading to $P(A^c) = 1 - P(A)$.

Generalization for Multiple Events

• For $n$ mutually exclusive events $A_1, A_2, ..., A_n$, the probability of their union is the sum of their individual probabilities:     $P(A_1 \cup A_2 \cup ... \cup A_n) = P(A_1) + P(A_2) + ... + P(A_n)$.

Practical Examples of Addition and Complement Rules

Example 2: Abstract Scenarios

• Scenario (a): $P(A) = 0.3, P(B) = 0.6, P(A \cap B) = 0.2$.

  • Probability of "at least one" ($A \cup B$):         $P(A \cup B) = 0.3 + 0.6 - 0.2 = 0.7$.

• Scenario (b): $P(A) = 0.2, P(B) = 0.7, P(A \cup B) = 0.8$.

  • Find probability of "both" ($A \cap B$):         $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.2 + 0.7 - 0.8 = 0.1$.

• Scenario (c): $P(A) = 0.3, P(B) = 0.6$, mutually exclusive.

  • $P(A \cup B) = 0.3 + 0.6 = 0.9$.

Example 3: System Components

• Scenario (a): System functions if at least one component functions. $P(A) = 0.4, P(B) = 0.5, P(A \cap B) = 0.1$.

  • $P(\text{System functions}) = P(A \cup B) = 0.4 + 0.5 - 0.1 = 0.8$.

  • $P(\text{System fails}) = P((A \cup B)^c) = 1 - 0.8 = 0.2$.

• Scenario (b): System functions if both function. $P(A) = 0.98, P(B) = 0.95, P(A \cup B) = 0.99$.

  • $P(\text{System functions}) = P(A \cap B) = 0.98 + 0.95 - 0.99 = 0.94$.

Example 4: University Course Percentages

• $40\%$ take Calculus ($A$), $30\%$ take Statistics ($B$), $10\%$ take both ($A \cap B$).

  • Percent taking at least one: $P(A \cup B) = 0.40 + 0.30 - 0.10 = 0.60$ (or $60\%$).

• $50\%$ Calculus ($A$), $20\%$ Statistics ($B$), $40\%$ at least one ($A \cup B$).

  • Percent taking both: $P(A \cap B) = 0.50 + 0.20 - 0.40 = 0.30$ (or $30\%$).

Probability with a Standard Deck of Cards

• Standard Deck Information: 52 cards total, 4 suits (Clubs, Spades - black; Diamonds, Hearts - red). Each suit has 13 cards: Ace, 2-10, and Face Cards (Jack, Queen, King). Total face cards = 12.

• Example 5 (Single card draw):

  • Spade ($A$) or King ($B$ ): $n(A) = 13, n(B) = 4, n(A \cap B) = 1$ (King of Spades).         $P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} \approx 0.3077$.

  • Heart ($C$) or Face Card ($D$): $n(C) = 13, n(D) = 12, n(C \cap D) = 3$ (Jack, Queen, King of Hearts).         $P(C \cup D) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} \approx 0.4231$.

Conditional Probability

• Definition: The probability of event $B$ occurring given that event $A$ has already occurred.

• Notation: $P(B|A)$, read as "probability of $B$ given $A$".

• General Formulas:

  • $P(B|A) = \frac{P(A \cap B)}{P(A)}$, provided P(A) > 0.

  • $P(A|B) = \frac{P(A \cap B)}{P(B)}$, provided P(B) > 0.

• Uniform Sample Space Formula: It is more convenient to use the ratio of outcomes:     $P(B|A) = \frac{n(A \cap B)}{n(A)}$.

• Concept - Reduced Sample Space: When calculating $P(B|A)$, the original sample space $S$ is effectively reduced to the outcomes that belong to event $A$.

Conditional Probability and Mutually Exclusive Events

• If $A$ and $B$ are mutually exclusive, then $P(A \cap B) = 0$.

• Therefore, $P(B|A) = 0$ and $P(A|B) = 0$. If one occur, it is impossible for the other to occur together with it.

Example 6: Conditional Contexts

• Two Children (Equally likely boy/girl):

  • Sample Space $S = \{bb, gg, gb, bg\}$, $n(S) = 4$.

  • Find $P(\text{both boys | at least one boy})$:         Event "at least one boy" $A = \{bb, gb, bg\}$, $n(A) = 3$.         Event "both boys" $B = \{bb\}$, $n(A \cap B) = 1$.         $P(B|A) = \frac{1}{3}$.

  • Find $P(\text{both boys | oldest is boy})$:         Event "oldest is boy" $C = \{bb, bg\}$, $n(C) = 2$.         $P(B|C) = \frac{1}{2}$.

• Tossing Single Coin Three Times:

  • Find $P(\text{3 heads | first toss was head})$:         $n(\text{First is head}) = 4$ outcomes; $n(\text{3 heads}) = 1$.         $P = \frac{1}{4}$.

  • Find $P(\text{3 heads | first two were heads})$:         $n(\text{First two heads}) = 2$ outcomes (hhh, hht); $n(\text{3 heads}) = 1$.         $P = \frac{1}{2}$.

Multiplication Rules

General Multiplication Rule

• Derived from the conditional probability formula.

• $P(A \cap B) = P(A) \times P(B|A)$

• $P(A \cap B) = P(B) \times P(A|B)$

• This rule calculates the probability that both events occur simultaneously.

Multi-Step Problem: Example 7 (Bus vs. Subway)

• Data: $P(\text{Bus}) = 0.70$, $P(\text{Subway}) = 0.30$.

• Conditional Probabilities: $P(\text{Late | Bus}) = 0.10$, $P(\text{Late | Subway}) = 0.05$.

• Results:

  • $P(\text{Bus and Late}) = 0.70 \times 0.10 = 0.0700$.

  • $P(\text{Subway and Late}) = 0.30 \times 0.05 = 0.0150$.

  • $P(\text{Bus and Not Late})$:         $P(\text{Not Late | Bus}) = 1 - 0.10 = 0.90$.         $P(\text{Bus and Not Late}) = 0.70 \times 0.90 = 0.6300$.

  • $P(\text{Subway and Not Late})$:         $P(\text{Not Late | Subway}) = 1 - 0.05 = 0.95$.         $P(\text{Subway and Not Late}) = 0.30 \times 0.95 = 0.2850$.

Independent Events

• Definition: Two events $A$ and $B$ are independent if the occurrence of one does not change the probability of the occurrence of the other.

• Formal Conditions:

  • $P(B|A) = P(B)$

  • $P(A|B) = P(A)$

• Special Multiplication Rule for Independent Events:

  • $P(A \cap B) = P(A) \times P(B)$.

• Example 10 (Retired Couple life expectancy):

  • $P(\text{Man lives 10 yrs}) = 0.75$, $P(\text{Woman lives 10 yrs}) = 0.85$.

  • Independent events.

  • $P(\text{Both alive}) = 0.75 \times 0.85 = 0.6375$.

  • $P(\text{Man dies and Woman lives}) = (1 - 0.75) \times 0.85 = 0.25 \times 0.85 = 0.2125$.

  • $P(\text{At least one alive}) = P(A \cup B) = 0.75 + 0.85 - 0.6375 = 0.9625$.

Comparison: Independent vs. Mutually Exclusive

• Exclusive Relationship:

  1. Independent events are never mutually exclusive (assuming P > 0). Because $P(A \cap B) = P(A)P(B) \neq 0$, whereas mutually exclusive events require the intersection to be 0.

  2. Mutually exclusive events are never independent. Knowing one occurs tells you the other cannot occur ($P(B|A) = 0$), which changes its probability (unless its original probability was already 0).

• Testing Independence (using Example 13a: Tornadoes and Flooding):

  • $P(T) = 0.15, P(F) = 0.08, P(T \cup F) = 0.21$.

  • $P(T \cap F) = 0.15 + 0.08 - 0.21 = 0.02$.

  • Verify $P(T \cap F) = P(T)P(F)$: $0.15 \times 0.08 = 0.012$.

  • Since $0.02 \neq 0.012$, the events are not independent.

  • Verify Mutual Exclusivity: Since $P(T \cap F) = 0.02 \neq 0$, they are not mutually exclusive.

Sequential Experiments

• Sequential Experiments: Experiments consisting of a sequence of two or more stages.

• Replacement Scenarios:

  • With Replacement: Trials are independent. Probability on stage 2 is the same as stage 1.

  • Without Replacement: Trials are dependent (not independent). Probability on stage 2 depends on outcomes of stage 1.

• Example 16 (Marbles without replacement):

  • Box: 2 Blue ($B$), 3 Red ($R$). Total 5.

  • $P(\text{Two Blue})$: $P(B_1) \times P(B_2 | B_1) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = 0.1000$.

  • $P(\text{Two Red})$: $P(R_1) \times P(R_2 | R_1) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = 0.3000$.

  • $P(\text{Different Colors})$:         $P((B_1 \cap R_2) \cup (R_1 \cap B_2)) = (\frac{2}{5} \times \frac{3}{4}) + (\frac{3}{5} \times \frac{2}{4}) = \frac{6}{20} + \frac{6}{20} = 0.6000$.

  • $P(\text{Same Color})$:         $P((B_1 \cap B_2) \cup (R_1 \cap R_2)) = 0.1000 + 0.3000 = 0.4000$.

Probability Tables

• Structure: Tables classifying items by two categorical variables. The main part (rows and columns) contains proportions of the sample falling into specific categories simultaneously.

• Joint Probabilities: The cells at the intersection of a row and column represent the probability of both categories occurring simultaneously ($P(A \cap B)$).

• Marginal Probabilities: The values in "Total" row/column represent probabilities of individual categories (e.g., $P(A)$, $P(S)$).

• Example 19 (Medical Students classification):

  • Variables: Smoking habits (Smoker $S$, Non-smoker $S^c$) and Alcohol consumption (Alcoholic $A$, Non-alcoholic $A^c$).

  • Data: $P(A \cap S) = 0.06$, $P(A \cap S^c) = 0.77$, $P(A^c \cap S) = 0.03$, $P(A^c \cap S^c) = 0.14$.

  • Total $P(S) = 0.09$; Total $P(A) = 0.83$.

  • $P(\text{Alcoholic | Smoker})$: $P(A|S) = \frac{P(A \cap S)}{P(S)} = \frac{0.06}{0.09} = 0.6667$.

  • $P(\text{Non-smoker | Alcoholic})$: $P(S^c|A) = \frac{P(A \cap S^c)}{P(A)} = \frac{0.77}{0.83} \approx 0.9277$.